Convex and Nonconvex Sweeping Processes with Velocity Constraints: Well-Posedness and Insights

Abstract

In this paper, we study some classes of sweeping processes with velocity constraints in the moving set. In addition to the solution existence and the solution uniqueness for the case of a moving convex constraint set, some results on the solution existence and the solution multiplicity where the constraint set is a finite union of disjoint convex sets are also obtained. Our main tool is a theorem on the solution sensitivity of parametric variational inequalities. Beside the traditional requirement that the constraint set moves continuously in the Hausdorff distance sense, we intensively use a new assumption on the local Lipschitz-likeness of the constraint set-valued mapping. The obtained results are compared with the existing ones and analyzed by several examples.

Appendix: Equivalent Norms

Claim 1

In the space $W^{1,\infty }((0,T),{\mathcal {H}})$ , the norm $\Vert f\Vert _{W^{1,\infty }}=\Vert f\Vert _{L^\infty }+\Vert f'\Vert _{L^\infty }$ is equivalent to the following one: $\Vert f \Vert =\Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert .$

Proof

Noting that $W^{1,\infty }((0,T),{\mathcal {H}})\subset W^{1,1}((0,T),{\mathcal {H}})$, one has f is an absolutely continuous function, which means

$$\begin{aligned} f(t)=f(0)+\int _0^t {\dot{f}}(s)ds,\quad \forall t\in [0,T]. \end{aligned}$$

From the proof of [16, Corollary 1.4.31] we can deduce that ${\dot{f}}(t)=f'(t)$ almost everywhere on (0, T). Then,

$$\begin{aligned} \Vert f\Vert _{W^{1,\infty }}=\Vert f\Vert _{L^\infty }+\Vert f'\Vert _{L^\infty }&= \Vert f\Vert _{L^\infty }+\Vert {\dot{f}}\Vert _{L^\infty } \\&= \mathop {\mathrm {ess\,sup }}\limits _{t\in (0,T)} \Vert f(t)\Vert + \Vert {\dot{f}}\Vert _{L^\infty }\\&= \sup _{t\in (0,T)} \Vert f(t)\Vert + \Vert {\dot{f}}\Vert _{L^\infty }\qquad (\Vert f(\cdot )\Vert \text { is continuous on} (0,T))\\&\ge \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert . \end{aligned}$$

(A.1)

We also have

$$\begin{aligned} \Vert f\Vert _{W^{1,\infty }}=\Vert f\Vert _{L^\infty }+\Vert f'\Vert _{L^\infty }&= \Vert f\Vert _{L^\infty }+\Vert {\dot{f}}\Vert _{L^\infty } \\&= \mathop {\mathrm {ess\,sup }}\limits _{t\in (0,T)} \Vert f(0)+\int _{0}^{t} {\dot{f}}(s)ds\Vert + \Vert {\dot{f}}\Vert _{L^\infty }\\&\le \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits _{t\in (0,T)} \Vert \int _{0}^{t} {\dot{f}}(s)ds\Vert + \Vert {\dot{f}}\Vert _{L^\infty }\\&\le \Vert f(0)\Vert + \int _{0}^{T} \Vert {\dot{f}}(s)\Vert ds\\&\quad + \Vert {\dot{f}}\Vert _{L^\infty } \qquad \text {(using Holder's inequality)}\\&\le \Vert f(0)\Vert +2\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert \\&\le 2(\Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert ). \end{aligned}$$

(A.2)

From (A.1) and (A.2), we obtain the desired result. $\square$

Claim 2

If M be a positive number, then the norm $\Vert \cdot \Vert _{W^{1,\infty }}$ on the space $W^{1,\infty }((0,T),{\mathcal {H}})$ is equivalent to the norm $\Vert \cdot \Vert _M$ defined by

$$\begin{aligned} \Vert f \Vert _M=\Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\left( e^{-Mt}\Vert {\dot{f}}(t)\Vert \right) ,\quad \forall f\in W^{1,\infty }((0,T),{\mathcal {H}}). \end{aligned}$$

Proof

As we have mentioned in the previous proof, ${\dot{f}} = f'$ almost everywhere on (0, T). From (A.1) it follows that

$$\begin{aligned} \begin{aligned} \Vert f\Vert _{W^{1,\infty }}&\ge \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert \\&\ge \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;(e^{-Mt}\Vert {\dot{f}}(t)\Vert )\qquad \text {(since }e^{-Mt}\le e^0=1\text {, for all }t\in [0,T]\text {)}. \end{aligned} \end{aligned}$$

(A.3)

By (A.2), we get

$$\begin{aligned} \begin{aligned} \Vert f\Vert _{W^{1,\infty }}&\le 2(\Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\Vert {\dot{f}}(t)\Vert )\\&=2\left[ \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\left( e^{Mt} e^{-Mt}\Vert {\dot{f}}(t)\Vert \right) \right] \\&\le 2\left[ \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\left( e^{MT} e^{-Mt}\Vert {\dot{f}}(t)\Vert \right) \right] \\&=2\left[ \Vert f(0)\Vert +e^{MT}\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\left( e^{-Mt}\Vert {\dot{f}}(t)\Vert \right) \right] \\&\le 2e^{MT}\left[ \Vert f(0)\Vert +\mathop {\mathrm {ess\,sup }}\limits \limits _{t\in (0,T)}\;\left( e^{-Mt}\Vert {\dot{f}}(t)\Vert \right) \right] . \end{aligned} \end{aligned}$$

(A.4)

Combining (A.3) and (A.4) yields the equivalence of the two norms under consideration. $\square$